9
$\begingroup$

Let $f$ be a differentiable function on $[0,1]$ and $a,b\in(0,1)$ such that $a<b$, $\int\limits_0^af(x)dx=\int\limits_b^1f(x)dx=0$. Show that: $$\left|\int_0^{1} f(x)\,dx\,\right|\leq\frac{1-a+b}{4}\,M$$ where $M=\sup\limits_{x\in[0,1]}|f'(x)|$.

$\endgroup$

closed as off-topic by Ben Sheller, Matthew Conroy, man and laptop, GNUSupporter 8964民主女神 地下教會, kimchi lover Feb 8 '18 at 0:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Ben Sheller, Matthew Conroy, man and laptop, GNUSupporter 8964民主女神 地下教會, kimchi lover
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ What are your ideas on this problem? $\endgroup$ – Dan Mar 2 '15 at 12:31
  • $\begingroup$ So in fact you're assuming as well the derivative is bounded on the unit interval? $\endgroup$ – Timbuc Mar 2 '15 at 12:54
  • $\begingroup$ @ Matthias - you're missing the condition on the integrals between $0$ and $a$, and $b$ and $1$. $\endgroup$ – Baron Mingus Mar 2 '15 at 13:00
  • $\begingroup$ @ Matthias $\int_0^a x \mathrm{d}x \neq 0$ for $a > 0$. $\endgroup$ – Baron Mingus Mar 2 '15 at 13:06
  • $\begingroup$ Upps I missed the last equality. Thx. $\endgroup$ – Matthias Mar 2 '15 at 13:07
7
$\begingroup$

Since $f$ is continuous, it must have zeroes in both intervals $[0,a]$ and $[b,1]$. By the Mean value theorem, we deduce that $$ |f(a)| \leq Ma,\qquad |f(b)| \leq M(1-b). $$

By the Mean value theorem again, there exists $c \in (a,b)$ such that $$\frac{1}{b-a}\left[\int_a^bf(x)dx - \frac{1}{2}(b-a) [f(a)+f(b)]\right] = -(c-\frac{a+b}{2})f'(c). $$ Therefore, $$ \left|\int_a^b f(x)dx\right| \leq \frac{b-a}{2} \left(|f(a)| + |f(b)| + (b-a)M\right) \leq \frac{b-a}{2}M. $$ The conclusion follows since $b-a < 1$ implies $\frac{b-a}{2} < \frac{b-a + 1}{4}$ and $$ \int_0^1 f(x)dx = \int_a^b f(x)dx. $$

$\endgroup$
  • 5
    $\begingroup$ It may help to note that in the second step he applied the mean value theorem to the function $g(x)=(\frac{a+b}{2}-x)f(x)+\int_0^x f(t) dt$. $\endgroup$ – J.R. Mar 2 '15 at 14:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.